Permutation group product Let $E,F$ be two vector spaces and $\varphi: \overbrace{E\times \cdots \times E}^{\text{$p$ times}} \to F$ a $p$-linear map. If $\sigma$ is a permutation on $S_{p}$, then we can define another $p$-linear map $\sigma \varphi: \overbrace{E\times \cdots \times E}^{\text{$p$ times}}\to F$ by:
$$(\sigma \varphi)(x_{1},...,x_{p}) := \varphi(x_{\sigma(1)},...,x_{\sigma(p)})$$
Now, my book says that if $\tau, \sigma$ are two permutations, then:
$$(\tau \sigma)\varphi = \tau(\sigma \varphi)$$
However, according to my calculations:
$$[(\tau \sigma)\varphi](x_{1},...,x_{p}) = \varphi(x_{\tau(\sigma(1))},...,x_{\tau(\sigma(p))}) = [\sigma(\tau \varphi)](x_{1},...,x_{p})$$
since $[\sigma(\tau\varphi)](x_{1},...,x_{p}) = (\tau\varphi)(x_{\sigma(1)},...,x_{\sigma(p)}) = \varphi(x_{\tau(\sigma(1))},...,x_{\tau(\sigma(p))})$. Where is my mistake?
 A: In general it is not true that $\varphi(x_{\tau(\sigma(1))},...,x_{\tau(\sigma(p))}) = [\sigma(\tau \varphi)](x_{1},...,x_{p})$ -- this only applies if $\tau$ and $\sigma$ commute in $S_p$. Take for example $p=3$, $\tau = (1\ 3), \sigma = (1\ 2\ 3)$. Then
$$\sigma\tau \phi = (1\ 2\ 3)(1\ 3)\phi = \phi(x_1,x_3,x_2) $$
$$\tau\sigma \phi = (1\ 3)(1\ 2\ 3)\phi = \phi(x_2, x_1, x_3)$$
What is true is that when evaluating $\sigma\tau\phi$ (for example), you can first apply $\tau$ to $\phi$ and then finally apply $\sigma$, or you can first apply $\sigma$ to $\tau$ and then apply this combined permutation ($\sigma\tau$) to $\phi$.
A: There is a mistake when you write
$$[(\tau \sigma)\varphi](x_{1},...,x_{p}) = \varphi(x_{\tau(\sigma(1))},...,x_{\tau(\sigma(p))}) = [\sigma(\tau \varphi)](x_{1},...,x_{p})$$
You have
$$\begin{aligned}(\tau \sigma)\varphi(x_{1},...,x_{p}) &= \varphi(x_{(\tau\sigma)(1)},...,x_{(\tau\sigma)(p))})\\
&= \varphi(x_{(\tau\circ \sigma)(1)},...,x_{(\tau\circ \sigma)(p))})\\
&= \varphi(x_{(\tau(\sigma(1))},...,x_{(\tau(\sigma(p))})\\
&= \tau(\varphi(x_{\sigma(1)},...,x_{\sigma(p)})\\
&= \tau(\sigma\varphi(x_1,...,x_p)\\
\end{aligned}$$
Therefore the equality of your book
$$(\tau \sigma)\varphi = \tau(\sigma \varphi)$$
